HCLTech Invitation  Walkin Drive in Noida,Chennai - 22nd June 2024  Multiple Domain



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IMS - Off Campus - 2024
Hexaware Technologies Limited · Mumbai,Pune,Chennai

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Good Sets


int solve(const string& s) {
    vector<int> lc;
    for (int i = 0; i < s.size(); ++i) {
        if (islower(s[i])) {
            lc.push_back(i);
        }
    }

    int m = 0;
    int c = 0;
    int p = -1;
    for (int i : lc) {
        if (p == -1 || s.substr(p + 1, i - p).compare(s.substr(p + 1, i - p)) != 0) {
            c += 1;
        } else {
            c = 1;
        }
        m = max(m, c);
        p = i;
    }

    return m;
}

#include <iostream>
#include <unordered_set>
#include <string>
using namespace std;
void generateSubstrings(const string &s, int len, unordered_set<string> &substrings) {
    for (int i = 0; i <= s.size() - len; ++i) {
        substrings.insert(s.substr(i, len));
    }
}

string findMinimalString(const string &s) {
    unordered_set<string> substrings;
    for (int len = 1; ; ++len) {
        substrings.clear();
        generateSubstrings(s, len, substrings);
        string candidate(len, 'a');
        while (true) {
            if (substrings.find(candidate) == substrings.end()) {
                return candidate;
            }
            int pos = len - 1;
            while (pos >= 0 && candidate[pos] == 'z') {
                candidate[pos] = 'a';
                --pos;
            }
            if (pos < 0) break;
            ++candidate[pos];
        }
    }
}

int main() {
    string S;
    cin >> S;
    cout << findMinimalString(S) << endl;
    return 0;
}. 

// Minimal String

Infosys ✅

#include <iostream>
#include <vector>
using namespace std;
vector<bool> sieve(int max_val) {
    vector<bool> is_prime(max_val + 1, true);
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i * i <= max_val; ++i) {
        if (is_prime[i]) {
            for (int j = i * i; j <= max_val; j += i) {
                is_prime[j] = false;
            }
        }
    }
    return is_prime;
}

int main() {
    int N;
    cin >> N;

    vector<int> A(N);
    int max_val = 0;
    for (int i = 0; i < N; ++i) {
        cin >> A[i];
        if (A[i] > max_val) {
            max_val = A[i];
        }
    }

    vector<bool> is_prime = sieve(max_val);

    int prime_count = 0, composite_count = 0;
    for (int i = 0; i < N; ++i) {
        if (is_prime[A[i]]) {
            prime_count++;
        } else {
            composite_count++;
        }
    }

    int good_pairs = prime_count * composite_count;
    cout << good_pairs << endl;

    return 0;
}.  

find good pairs in array
Infosys ✅